# doubt from capacitance and capacitors

Can a charged capacitor mathematically be represented as uncharged capacitor connected in series with battery of opposing emf ?

• It can be done to find the heat generated but not to find final charges – user1825567 Apr 21 '16 at 7:58
• I think the argument should be "Can a charged capacitor mathematically be represented as a battery of opposing emf ?" – jim Apr 21 '16 at 10:55

The two networks will be equivalent if the e.m.f in the second network is equal to $Q/C$ in the first. Then you will get two networks with same initial voltage across their terminals and same impedance too (since $z_{voltage source}=0$), so they are equivalent.
Applying KVL at an instant $t$ until which a charge $q$ has passed through the circuit, in the first one we get:$$\dfrac{Q-q}{C}=Ri$$
which is a differential equation in $q$.
In the second:$$E=Ri+\dfrac{q}{C}$$but $E=\dfrac{Q}{C}$, then:$$\dfrac{Q-q}{C}=Ri$$ so it is the same differential equation, and will give the same equation for in both circuits.